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Ta thay S co 50 so hang ma
\(\frac{1}{50}>\frac{1}{100},\frac{1}{51}>\frac{1}{100},\frac{1}{52}>\frac{1}{100},...,\frac{1}{99}>\frac{1}{100}\)
=> cong tung ve 50 bdt cung chieu ta duoc
\(S>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{50}{100}=\frac{1}{2}\) (do S co 50 so hang )
Vay S>1/2 dpcm
a) \(\frac{31}{23}-\left(\frac{7}{23}+\frac{8}{23}\right)\)
\(=\frac{31}{23}-\frac{15}{23}\)
\(=\frac{16}{23}\)
b) \(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
\(=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)
\(=\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)
\(=\frac{1}{3}+\frac{-67}{67}+\frac{41}{41}\)
\(=\frac{1}{3}-1+1\)
\(=\frac{1}{3}\)
c) \(\frac{38}{45}-\left(\frac{8}{45}-\frac{17}{52}-\frac{3}{11}\right)\)
\(=\frac{38}{45}-\frac{8}{45}+\frac{17}{52}+\frac{3}{11}\)
\(=\frac{30}{45}+\frac{17}{52}+\frac{3}{11}\)
\(=\frac{2}{3}+\frac{17}{52}+\frac{3}{11}\)
\(=\frac{104+51}{156}+\frac{3}{11}\)
\(=\frac{155}{156}+\frac{3}{11}\)
\(=\frac{156}{156}-\frac{1}{156}+\frac{3}{11}\)
\(=1-\frac{1}{156}+\frac{3}{11}\)
\(=1-\left(\frac{11-468}{1716}\right)\)
\(=1-\frac{-457}{1716}\)
\(=1+\frac{457}{1716}\)
\(=\frac{2173}{1716}\)
a)31/23-(7/32+8/23)=31/23-7/32-8/23=(31/23-8/23)-7/32=1-7/32=25/32
a)\(=\frac{-3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}\)
\(=\left(\frac{-3}{7}+\frac{3}{7}\right)-\left(\frac{15}{26}+\frac{2}{13}\right)\)
\(=0-\frac{19}{26}\)
\(=-\frac{19}{26}\)
c)\(=\frac{-11}{23}.\left(\frac{6}{7}+\frac{8}{7}\right)-\frac{1}{23}\)
\(=\frac{-11}{23}.2-\frac{1}{23}\)
\(=\frac{-22}{23}-\frac{1}{23}\)
\(=-1\)
Ta có :
\(\frac{1}{50}>\frac{1}{100}\)
\(\frac{1}{51}>\frac{1}{100}\)
\(\frac{1}{52}>\frac{1}{100}\)
\(............\)
\(\frac{1}{98}>\frac{1}{100}\)
\(\frac{1}{99}>\frac{1}{100}\)
\(\Rightarrow\)\(S=\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{98}+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}+\frac{1}{100}\)
Do từ \(50\) đến \(99\) có \(99-50+1=50\) số nên có \(50\) phân số \(\frac{1}{100}\)
Suy ra :
\(S>50.\frac{1}{100}=\frac{50}{100}=\frac{1}{2}\)
Vậy \(S>\frac{1}{2}\)
Chúc bạn học tốt ~
Mình nhầm chứng tỏ tổng của các phân số sau đây lớn hơn \(\frac{1}{2}\)